Proof of a conjectured formula for A208531 from Toshitaka Suzuki (suzuki(AT)scio.co.jp) July 29 2019 Theorem: A208531(n) = (2*n^3+3*n^2+n-366)/6 for n > 8. Proof: Define: issum(m, n) = 1 if m is the sum of distinct squares of 1,4,9,…,n^2, 0 otherwise. By computing all combinations, issum(k, 9) = 1 for 0 <= k <= 128 except A001422. And computing all combinations, issum(k, 10) = 1 for 129 <= k <= 256. Then issum(k, 11) = 1 for 129+121 <= k <= 256+121 by adding 11^2=121, so issum(k, 11) = 1 for 129 <= k <= 377. Similarly, issum(k, 12) = 1 for 129+144 <= k <= 377+144 so issum(k, 12) = 1 for 129 <= k <= 521 issum(k, 13) = 1 for 129+169 <= k <= 521+169 so issum(k, 13) = 1 for 129 <= k <= 690 ... issum(k, n) = 1 for 129 <= k <= 256 + Sum_{m=11..n}m^2 # Because # (256+Sum_{m=11..n}m^2)-129+1 - (n+1)^2 # = (2(n-10)^3+57(n-10)^2+529(n-10)+42)/6 > 0 256 + Sum_{m=11..n}m^2 > (Sum_{m=1..n}m^2) / 2 for n >= 10 and issum(m, n) = issum((Sum_{m=1..n}m^2) - m, n) (by reverse use or not) so issum(k, n) = 1 for 0 <= k <= Sum_{m=1..n}m^2 except A001422 and (Sum_{m=1..n}m^2) - A001422 for n >= 10. Therefore A208531(n) = (Sum_{m=1..n}m^2) + 1 - 31*2 = (2*n^3+3*n^2+n-366)/6 for n >= 10 (31 is not in A208531 ) and A208531(9)=(2*9^3+3*9^2+9-366)/6, hence A208531(n) = (2*n^3+3*n^2+n-366)/6 for n > 8. QED